What Forces Act on a Mass Sliding Inside a Hoop?

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A mass of 0.52 kg slides inside a frictionless hoop with a radius of 1.10 m and a speed of 4.25 m/s at the top. The total energy at the top is calculated to be 15.9189 J, which is used to determine the speed at a 44° angle, yielding a velocity of 7.44503 m/s. The centripetal force is initially calculated as 26.2026 N, but adjustments for the gravitational component are necessary. The net force exerted on the mass includes both the centripetal force and the gravitational force's radial component. Properly accounting for these forces is essential to find the correct force with which the mass pushes on the hoop.
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Homework Statement


A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 4.25 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 44.0°.


Homework Equations


E = Pe + Ke
Pe = m * g * h
Ke = 1/2 * m * v^2
F=mv^2/r


The Attempt at a Solution



Alright so energy at the top = Pe + Ke
So m*g*h + 1/2*m*v^2 = 11.2226 + 4.69625 = 15.9189

The energy in the bottom is going to be equal to the energy at the top
The height at the bottom is R-Rcos(44) = .295511
15.9189 = .52*9.81*.29511 + .5*.52*v^2
V=7.44503

Centripital force = mv^2/r, so .52*7.44503^2/1.1 = 26.2026 N

But this is wrong. Now I thought it might be reduced due to gravity.
so F=26.2026-m*g*cos(44) = 22.5331
Neither answer works.
Help?
 
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Staerke said:

Homework Statement


A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 4.25 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 44.0°.

Homework Equations


E = Pe + Ke
Pe = m * g * h
Ke = 1/2 * m * v^2
F=mv^2/r

The Attempt at a Solution



Alright so energy at the top = Pe + Ke
So m*g*h + 1/2*m*v^2 = 11.2226 + 4.69625 = 15.9189

The energy in the bottom is going to be equal to the energy at the top
The height at the bottom is R-Rcos(44) = .295511
15.9189 = .52*9.81*.29511 + .5*.52*v^2
V=7.44503

Centripetal force = mv^2/r, so .52*7.44503^2/1.1 = 26.2026 N

But this is wrong. Now I thought it might be reduced due to gravity.
so F=26.2026-m*g*cos(44) = 22.5331
Neither answer works.
Help?
Sorry that nobody responded sooner.

What is the angle 44.0° in relation to.

Your answer looks right for centripetal force, assuming the velocity is right.

The radial component of the net force exerted on M is equal to Mv2/r, the centripetal force. However, there are two contributions to this force.
One is the force exerted on M by the hoop.

The other is gravity. There a component of the gravitational force which is in the radial direction. This is what you haven't taken into account.​
 
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